The easiest way to find the turning point is when the quadric is in it’s turning form y = a(x-h)^2 +k, where (h,k) is the turning point.
Let’s put the function in the following form:
y = 2x^2 +5x -2 = 2(x^2 +\frac{5}{2}x -1) = 2(x^2 +2\frac{5}{4}x +(\frac{5}{4})^2 -(\frac{5}{4})^2 -1)=
= 2((x +\frac{5}{4}) ^2 -(\frac{5}{4})^2 -1) = 2(x +\frac{5}{4}) ^2 - 2\cdot (\frac{25}{16}+1) = 2(x +\frac{5}{4}) ^2 - 2\cdot \frac{41}{16} =
= 2(x +\frac{5}{4}) ^2 -\frac{41}{8}
\to y =2(x -(-\frac{5}{4})) ^2 + (-\frac{41}{8})
Comparing this with the turning form: y = a(x-h)^2 +k, we can see that h = -\frac{5}{4} and k = -\frac{41}{8}.
From this we conclude that the turning point is: (h,k) = ( -\frac{5}{4},-\frac{41}{8}) . So the correct solution is (B)